Javascript regular expression to extract characters from mid string with optional end character












2















I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.



Example #1:



"rilaS=testingabc"


should extract:



"testingabc"


Example #2:



"rilaS=testing123&thistest"


should extract:



"testing123"


This is what I have so far (Javascript):



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=(.*)[&]{0,1}");

var newStr = tmpStr[1];

alert(newStr);


But it does not detect that the end should be the ampersand (if found). Thank you before hand.



Answer (By ggorlen)



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=([^&]*)");

var newStr = tmpStr[1];

alert(newStr);





javascript regex







share|improve this question

























  • Str.match("S=(.*)[&]"); is not enought ?

    – V. Sambor
    Nov 27 '18 at 16:37
















2















I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.



Example #1:



"rilaS=testingabc"


should extract:



"testingabc"


Example #2:



"rilaS=testing123&thistest"


should extract:



"testing123"


This is what I have so far (Javascript):



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=(.*)[&]{0,1}");

var newStr = tmpStr[1];

alert(newStr);


But it does not detect that the end should be the ampersand (if found). Thank you before hand.



Answer (By ggorlen)



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=([^&]*)");

var newStr = tmpStr[1];

alert(newStr);





javascript regex







share|improve this question

























  • Str.match("S=(.*)[&]"); is not enought ?

    – V. Sambor
    Nov 27 '18 at 16:37














2












2








2








I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.



Example #1:



"rilaS=testingabc"


should extract:



"testingabc"


Example #2:



"rilaS=testing123&thistest"


should extract:



"testing123"


This is what I have so far (Javascript):



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=(.*)[&]{0,1}");

var newStr = tmpStr[1];

alert(newStr);


But it does not detect that the end should be the ampersand (if found). Thank you before hand.



Answer (By ggorlen)



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=([^&]*)");

var newStr = tmpStr[1];

alert(newStr);





javascript regex







share|improve this question
















I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.



Example #1:



"rilaS=testingabc"


should extract:



"testingabc"


Example #2:



"rilaS=testing123&thistest"


should extract:



"testing123"


This is what I have so far (Javascript):



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=(.*)[&]{0,1}");

var newStr = tmpStr[1];

alert(newStr);


But it does not detect that the end should be the ampersand (if found). Thank you before hand.



Answer (By ggorlen)



var Str = "rilaS=testing123&thistest";

var tmpStr = Str.match("S=([^&]*)");

var newStr = tmpStr[1];

alert(newStr);





javascript regex




javascript regex






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 27 '18 at 16:49







Robert Smith

















asked Nov 27 '18 at 16:27









Robert SmithRobert Smith

303213




303213













  • Str.match("S=(.*)[&]"); is not enought ?

    – V. Sambor
    Nov 27 '18 at 16:37



















  • Str.match("S=(.*)[&]"); is not enought ?

    – V. Sambor
    Nov 27 '18 at 16:37

















Str.match("S=(.*)[&]"); is not enought ?

– V. Sambor
Nov 27 '18 at 16:37





Str.match("S=(.*)[&]"); is not enought ?

– V. Sambor
Nov 27 '18 at 16:37












2 Answers
2






active

oldest

votes


















4














You may use /S=([^&]*)/ to grab from an S= to end of line or &:






["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>

  console.log(s.match(/S=([^&]*)/)[1])

);








share|improve this answer





















  • 1





    Thank you worked perfect!

    – Robert Smith
    Nov 27 '18 at 16:48



















0














Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.






share|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    You may use /S=([^&]*)/ to grab from an S= to end of line or &:






    ["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
    
  console.log(s.match(/S=([^&]*)/)[1])
    
);








    share|improve this answer





















    • 1





      Thank you worked perfect!

      – Robert Smith
      Nov 27 '18 at 16:48
















    4














    You may use /S=([^&]*)/ to grab from an S= to end of line or &:






    ["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
    
  console.log(s.match(/S=([^&]*)/)[1])
    
);








    share|improve this answer





















    • 1





      Thank you worked perfect!

      – Robert Smith
      Nov 27 '18 at 16:48














    4












    4








    4







    You may use /S=([^&]*)/ to grab from an S= to end of line or &:






    ["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
    
  console.log(s.match(/S=([^&]*)/)[1])
    
);








    share|improve this answer















    You may use /S=([^&]*)/ to grab from an S= to end of line or &:






    ["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
    
  console.log(s.match(/S=([^&]*)/)[1])
    
);








    ["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
    
  console.log(s.match(/S=([^&]*)/)[1])
    
);





    ["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s =>
    
  console.log(s.match(/S=([^&]*)/)[1])
    
);






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 27 '18 at 16:41

























    answered Nov 27 '18 at 16:34









    ggorlenggorlen

    7,1883826




    7,1883826








    • 1





      Thank you worked perfect!

      – Robert Smith
      Nov 27 '18 at 16:48














    • 1





      Thank you worked perfect!

      – Robert Smith
      Nov 27 '18 at 16:48








    1




    1





    Thank you worked perfect!

    – Robert Smith
    Nov 27 '18 at 16:48





    Thank you worked perfect!

    – Robert Smith
    Nov 27 '18 at 16:48













    0














    Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.






    share|improve this answer




























      0














      Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.






      share|improve this answer


























        0












        0








        0







        Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.






        share|improve this answer













        Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 27 '18 at 16:57









        TroffTroff

        101




        101






























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